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2 years ago

Porque deja de funcionar una estufa electrica cuando la electricidad es excesiva??

Physics

2 answers:

Brums [2.3K]2 years ago

7 0

<h3>Porque deja de funcionar una estufa electrica cuando la electricidad es excesiva?</h3>

Es normal que los quemadores superiores de una estufa eléctrica o una estufa se enciendan y apaguen en configuraciones distintas a Hi. El quemador se encenderá y apagará más de lo normal cuando se utilizan cacerolas que no son planas o que son del tamaño incorrecto para el quemador.

<h2>☆彡Hanna</h2>

#CarryOnLearning

Artyom0805 [142]2 years ago

5 0

<h3>Porque deja de funcionar una estufa electrica cuando la electricidad es excesiva?</h3>

Es normal que los quemadores superiores de una estufa eléctrica o una estufa se enciendan y apaguen en configuraciones distintas a Hi. El quemador se encenderá y apagará más de lo normal cuando se utilizan cacerolas que no son planas o que son del tamaño incorrecto para el quemador.

<h2>☆彡Hanna</h2>

#CarryOnLearning

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An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.5 kg mass of the pulley is concentrated on i

alexandr1967 [171]

Explanation:

63 kg ice skater finishes her performance and crossed the finish line with a speed of 10.8 m/s

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3 years ago

What is the velocity of a wave with a frequency of 45 Hertz and a wavelength of 3 meters?

Juli2301 [7.4K]

Explanation:

By using v=( f )x( lambda )

v= 45 s^-1 x 3 m

Therefore v = 135 ms^-1

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2 years ago

What’s the potential difference across a 5.0 ohms resistor that carries a current of 5.0A

vivado [14]

Answer:

The correct answer is B-25 V

Explanation:

We apply Ohm's Law, according to which:

V = i x R

V = 5A x 5Ω

V= 25 V

Being V the potential difference whose unit is the VOLT, i the current intensity (Ampere) and R the electrical resistance (ohm)

3 0

3 years ago

A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross

Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

$F_D=\frac{1}{2}\rhoC_dAV^2$

Our values are:

$V=30.1m/s$

$C_d=0.45$

$A=3.3m^2$

$\rho=1.2kg/m^3$

Replacing,

$F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2$

$F_D=807.25N$

We need calculate now the energy lost through a time T, then,

$W = F_D d$

But we know that d is equal to

$d=vt$

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

$W=(807.25N)(30.1m/s)(3600s/1hr)$

$W=87.47*10^6J$ (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0

2 years ago

PLS SHOW WORK !!!!!!! ASAP

Lynna [10]

Answer:

1- For the track B. The potential energy is the same for the two cars, but because of the slope of the track, the car B earn kinetic energy faster. The gravitation acceleration of the cars will be g•sinθ, and the angle of the track B will have a bigger value for sinθ

2- The conservation of energy applies because the roller coaster is a closed track. When a car climb the track, it earn GPE, which is given by mgh, when it get down in the track, it transform GPE in KE, which is given in 1/2mv².

Position of car (m) GPE KE GPE + KE

top (30m) 60000 0 60000

bottom (0m) 0 60000 60000

halfway down (15m) 30000 30000 60000

three-quarters way down 15000 45000 60000

4 0

2 years ago